I'm having a tough time figuring out the 'correct' normalization for extending absolute values of local fields. I'm also trying to piece together how this interacts with the global theory, so below is essentially a discussion of my thoughts and a few questions at the end. Am I thinking about this the right way?

If $K/\mathbb Q$ is finite then every nonzero ideal $\mathfrak a\subseteq \mathcal O_K$ has a unique factorization $$ \mathfrak a=\prod_{\mathfrak p: \,\text{prime}}\mathfrak p^{e_\mathfrak p(\mathfrak a)}. $$ The function $v_\mathfrak p:\mathcal O_K\rightarrow \mathbb Z\cup \infty$ defined by $v_\mathfrak p(x)={e_\mathfrak p(x\mathcal O_K)}$ defines a discrete valuation on $\mathcal O_K$. We can extend this to a valuation on $K$ by noting that all elements of $K$ can be represented by a quotient of algebraic integers. Fixing a prime $p$ and a prime ideal $\mathfrak p\mid p$, for any $0<a<1$ we have an induced nonarchimedean absolute value on $K$, given by $|x|_\mathfrak p=a^{v_\mathfrak p(x)}$. Completing $K$ with respect to this absolute value results in a finite extension $K_\mathfrak p$ of $\mathbb Q_p$.

If $e$ and $f$ are the ramification and inertia degrees of $p$, by Neukirch Theorem 4.8, I also understand that $K_\mathfrak p$ has a unique absolute value extending the $p$-adic absolute value $|\cdot |_p$ on $\mathbb Q_p$ , given by $$ |x|_\mathfrak p=|N_{K_\mathfrak p/\mathbb Q_p}(x)|_p^{\frac{1}{ef}} $$ since $[K_\mathfrak p:\mathbb Q_p]=ef$. This forces us to fix a value for $a$ above: since we want the absolute value on $K_\mathfrak p$ to extend $|\cdot |_p$, we'd like to have $|p |_\mathfrak p=p^{-1}$. Therefore, we need $ a^{v_\mathfrak p(p)}=|N_{K_\mathfrak p/\mathbb Q_p}(p)|^{\frac{1}{ef}}, $ which, since $N_{K_\mathfrak p/\mathbb Q_p}(p)=p^{ef}$ (do I need to assume the extension is Galois here?) and $v_\mathfrak p(p)=e$, implies that $a=p^{-1/e}$. Therefore, we normalize so that $$ |x|_\mathfrak p=p^{-v_\mathfrak p(x)/e} $$ is the absolute value extending that of $\mathbb Q_p$. All this said, Serre mentions in page 27 of Local Fields, that for a locally compact field (which is the case for $K_\mathfrak p$), there is a "canonical way to choose the number $a$ [defined the same as I have done above]: one takes $a=q^{-1}$, where $q$ is the number of elements in the reside field". In our case, I think that $q=|\mathcal O_{K_\mathfrak p}/\mathfrak p|=p^f$? If so, then it seems like Serre would have us take $a=p^{-f}$... What is the advantage of doing this? Am I making some horrible mistake here? Is there a better way to think about (normalized) valuations on extensions? Is there a way to 'get around' computing the field norm whenever we'd like to take the absolute value of an an element?

I realize I've bunched a few questions together here, and I'd be happy to write them separately if someone with more experience on this site would recommend it. That said, partial answers or even just comments would be welcome.

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    $\begingroup$ I think that the “canonical way” is what makes the Product Formula work for a global field $K$. But it does not extend the canonical $p$-abs. value of $\Bbb Q$. $\endgroup$
    – Lubin
    Feb 15, 2019 at 5:49
  • $\begingroup$ The product formula for number fields says that $\prod_l |\sigma_l(\alpha)|_\infty = |N_{K/\mathbb{Q}}(\alpha)|_\infty = N(\alpha O_K)= \prod_{\mathfrak{p} | (\alpha)} N(\mathfrak{p})^{v_\mathfrak{p}(\alpha)}$ so you'd take $|\alpha|_\mathfrak{p} = N(\mathfrak{p})^{-v_\mathfrak{p}(\alpha)} = p^{-f_\mathfrak{p}v_\mathfrak{p}(\alpha)}$ and make the complex absolute values appear twice to obtain $\prod_v |\alpha|_v = 1$, no ramification index here $\endgroup$
    – reuns
    Feb 16, 2019 at 19:30

1 Answer 1


You take a "local field" as a finite extension of a $\mathbf Q_p$, but my discussion below will concern any field which is locally compact w.r.t. a non archimedean absolute value , i.e. which is complete w.r.t. a discrete valuation and has finite residue field.

1) As far as $K$ with an absolute value is considered as solely a normed space, only the topology of $K$ matters. Recall that two absolute values ${\mid . \mid}_1$ and ${\mid . \mid}_2$ are equivalent iff they induce homeomorphic topologies on $K$, iff there exists a strictly positive real $c$ s.t. ${\mid . \mid}_1={{\mid . \mid}_2}^c$. An equivalence class of absolute values is called a place. The "normalization" suggested by Serre, as you recalled it, for a field $K$ with a discrete valuation ${\mid\mid . \mid\mid}$ and finite residue field of cardinal $q$, consists in putting ${\mid\mid \pi \mid\mid}=q^{-1}$ for a uniformizer $\pi$. It is "canonical" in the following sense. If $K$ is locally compact, introduce the Haar measure $\mu(.)$ on $K^+$ which is uniquely determined by the condition $\mu(O_K)=1$. Then ${\mid\mid . \mid\mid}$ is the unique absolute value which verifies $\mid\mid y\mid\mid = \mu(x+yO_K)$ (because of the invariance of $\mu$, the RHS is independent of $x$).

2) If we consider a finite extension $L/K$ of degree $n$, a given absolute value ${\mid . \mid}_K$ admits exactly one extension to an absolute value ${\mid . \mid}_L$, which is defined by ${\mid x \mid}_L= {{\mid N(x) \mid}_K}^{1/n}$, where $N$ is the norm of $L/K$. Here a discrepancy appears with the previous normalized absolute values, in that ${\mid\mid x\mid\mid}_L= {\mid\mid N(x)\mid\mid}_K$. In some sense, this new relation is more natural w.r.t. to a given $x$, because its formulation behaves coherently with a change of extension $L/K$.

3) Let us now consider the global-local relations between a global field $F$ and its completed local fields $F_v$ for all places $v$ (including the archimedean ones). For a given $F$, a fundamental result is the so called product formula : $\prod_v {\mid\mid x\mid\mid}_v=1$ for $x\in K^*$, in which the normalized archimedean absolute values are by definition the usual absolute value in $\mathbf R$ or the square of the usual modulus in $\mathbf C$. Local CFT is expressed in terms of the multiplicative groups of a local field $K$ and its extensions. Modern CFT for a global field $F$ is expressed in terms of the idèle group $J_F$ and the idèle class group $C_F=J_F/F^*$ (I don't recall the definitions). The topology on $J_F$ is not the product toplogy, but the so called "restricted topological product" of the ${F_v}^*$ w.r.t. the units $U_v$ inside. The subgroup $F^*$ is discrete. The (huge) connected component of $1$ in $C_F$ plays an annoying important role ./.

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    $\begingroup$ So in $ \hat{O_K}$ then $\prod_{v \ fin} |x|_v = \mu(a+x\hat{O_K})$ is the Haar measure. But when looking at the whole adeles it is less obvious why we'd want $\prod_v |x|_v = 1,x \in K^*$ and why the complex absolute values appear twice, beside CFT and characters of $\Bbb{A_K}^\times/K^\times$ $\endgroup$
    – reuns
    Feb 16, 2019 at 20:24
  • $\begingroup$ No, it's the product over all places of the global field, and it's $1$. I didn't say that it's equal to the Haar mesure, which is a local thing. What is $\hat O_K$ ? $\endgroup$ Feb 16, 2019 at 21:57
  • $\begingroup$ $K$ a number field $\hat{O_K} = \varprojlim O_K/I = \prod_\mathfrak{p} \varprojlim O_K/\mathfrak{p}^n$ the ring of integers of the finite adeles. The complex absolute values appear twice in $\prod_v |x|_v=1$ because they appear twice in the LHS of $\prod_j |\sigma_j(\alpha)| = |N_{K/\mathbb{Q}}(\alpha)| = N(\alpha O_K)= \prod_\mathfrak{p} N(\mathfrak{p})^{v_\mathfrak{p}(\alpha)} = \prod_{v \ fin} |\alpha|_v^{-1}$ $\endgroup$
    – reuns
    Feb 16, 2019 at 22:14
  • $\begingroup$ Ah, OK. I added an explanation about the formula involving the Haar measure. $\endgroup$ Feb 16, 2019 at 22:33

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